A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find

By BYJU'S Exam Prep

Updated on: September 25th, 2023

(a) the total distance moved by train,

(b) the maximum speed attained by the train, and

(c) the position(s) of the train at half the maximum speed.

(a) the total distance moved by train is 2.7 km,

(b) the maximum speed attained by the train is 60 m/s, and

Step 1: Given that acceleration, a = 2.0 m/s²

Initial velocity, u = 0

Step 2:

Using the equations of motion we have to calculate the final velocity and the total distance traveled.

Final velocity, v = u + at

Distance traveled, x = ut + ½ at²

Step 3: In case of accelerated motion

Given that, a = 2.0 m/s²

Let v₁ represent the final speed before applying breaks.

we know that, initial velocity, u₁ = 0

Time taken, t₁ = ½ x 60 = 30 seconds

Final velocity is now provided by,

v1 = u1 + at1

Substituting the values we get

v1 = 0 + 2 x 30

v1 = 60 m/s

The top speed reached after applying brakes is 60 m/s²

Distance traveled

x = u₁t₁ + ½ at₁²

x = 0 + ½ x 2 x 30²

x = 900 m

x = 0.9 km

The maximum speed attained is 60 m/s², Let y be the distance covered while moving at half the maximum speed of 30 m/s.

From the equation now

v₁² = u₁² + 2ax

30² = 0⁰ + 2 x 2 x y

y = 900/4 = 225 m ….. (1)

As a result, the starting point is 225 metres away.

Table of content

1. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute.

Step 4: In case of decelerated motion

Final velocity, v₂ = 0 m/s

Initial velocity, u₂ = 60 m/s

Time taken, t₂ = 60 s

From the equation

v₂ = u₂ + at₂

a = (0-60)/60 = -1 m/s²

Distance traveled,

x = u₂t₂ + ½ at₂²

Substituting the values we get

x = 60 x 60 + ½ x (-1) x 60²

x = 3600 – 1800

x = 1800 m

x = 1.8 km

The total distance covered = 0.9 + 1.8 = 2.7 km

Let z represent the distance travelled in this portion at half the speed.

From

v₂² = u₂² + 2az

30² = 60²+ 2 x (-1) x z

2z = 60² – 30²

2z = 2700

z = 1350m

z = 1.35 km …. (2)

Step 5: Conclusion

(a) Total distance traveled by train is given as,

s = x + z

= 900 + 1800

= 2700 m

= 2.7 km

(b) The train can travel at a maximum speed of 60 m/s.

Two places will operate at half the maximum speed (From the point of start),

The position is 225 metres from the starting point.

After braking, the position is = 1.35 km + 0.9 km = 2.25 km

Summary:-

A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute.

(a) the total distance moved by train is 2.7 km,

(b) the maximum speed attained by the train is 60 m/s, and

Related Questions:-

A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find

A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute.

Recent Posts

A train starts from rest and moves with a constant acceleration of 2.0 m/s² for half a minute. The brakes are then applied and the train comes to rest in one minute.